# g - BUSI 403 Problem Set G Solution Fall 2008 Due: 11/11/08...

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Unformatted text preview: BUSI 403 Problem Set G Solution Fall 2008 Due: 11/11/08 Problem 1 1. Control limits and the center lines for the X and R charts. Bulb Life (1000 hours) Sample Number Mean () Range (R) 1 2 3 4 5 1 4.291 4.271 4.112 4.263 4.210 4.229 0.179 2 4.289 4.190 4.164 4.232 4.264 4.228 0.125 3 4.222 4.146 4.106 4.292 4.174 4.188 0.186 4 4.196 4.191 4.158 4.285 4.159 4.198 0.127 5 4.175 4.283 4.259 4.198 4.200 4.223 0.108 6 4.157 4.255 4.222 4.253 4.240 4.225 0.098 7 4.248 4.269 4.149 4.131 4.235 4.206 0.138 Using X and R values, we get X=4.214 and R=0.137. And the factors can be read from the slides given in class for sample size=5: Factors A2 0.58 D3 0.00 D4 2.11 Then the following line and limits can be calculated: X bar chart Center Line (CL) R chart Upper Control Limit (UCL) X A 4.294 Lower Control Limit (LCL) X A 4.134 1 R=0.137 X=4.214 RD 0.290 RD 0 BUSI 403 Problem Set G Solution Fall 2008 Due: 11/11/08 2. i. Plot X and R charts Mean Chart (X-Bar) 4.350 4.300 4.250 4.200 4.150 4.100 1 2 3 4 5 Sample Mean 6 CL 7 8 UCL 9 10 9 10 LCL Mean Chart (R-Bar) 0.400 0.350 0.300 0.250 0.200 0.150 0.100 0.050 0.000 ‐0.050 1 2 3 4 5 Sample Range CL 6 7 UCL 8 LCL ii. Is the production process still in control? Why or why not? Although all observations are between the UCL and LCL of the X chart, several points are above the UCL of R chart. Therefore the process is no longer in control. 2 BUSI 403 Problem Set G Solution Fall 2008 Due: 11/11/08 Problem 2 a. Control Limits 1000 15 Sample size n # of samples Week 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Total Average # of defectives per lot 18 22 10 11 15 26 19 12 25 9 17 15 21 19 24 263 17.533 Proportion 0.02 0.02 0.01 0.01 0.02 0.03 0.02 0.01 0.03 0.01 0.02 0.02 0.02 0.02 0.02 0.263 0.01753 From the above table, p = 0.01753 σ 0.00415 p chart : = p = 0. 01753 = p 3 σP = 0. 01753+ 3 * 0.00415 = 0.03 =max(0, p 3 σP ) = max(0,0. 01753- 3 * 0.00415) = 0.00508 Center Line (CL) Upper Control Limit (UCL) Lower Control Limit (LCL) b. Process Control p-Chart Fraction Defective (p) 0.035 0.03 0.025 0.02 0.015 0.01 0.005 0 1 2 3 4 5 Proportion (p) CL 6 7 UCL 8 9 10 LCL The process is no longer in control because 2 data points have exceeded the upper control limit. 3 BUSI 403 Problem Set G Solution Fall 2008 Due: 11/11/08 Problem 3 a. Cp measure USL 90 LSL 80 Facility A Facility B μ 82 86 σ 1 1.2 Cp USL LSL USL LSL 1.67 σ 1.39 . σ Since both Cp are greater than 1, both facilities are capable. b. Cpk measure USL 90 LSL 80 Facility A Facility B μ 82 86 σ 1 1.2 min Cp = min USL μ μ LSL σ , min σ 90 82 82 80 , = 0.67 31 31 = min USL μ μ LSL σ σ 90 86 86 80 , = 1.11 3 1.2 3 1.2 Since only Cpk for facility B is greater than 1, only facility B is capable. 4 , ...
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## This document was uploaded on 11/04/2011 for the course BUSI 403 at UNC.

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